awk: Extracting part of the buffer

Shell Programming and Scripting
1

Hi,

I am trying to extract part of a line using "awk". My requirement is to extract the value $6 (which is the last parameter) from a line. As the sixth value contains some space, i am getting only part of the string. so i am trying to extract from $6 to the end of the buffer.
How to do it using awk?

Plz help.

2

Choose the appropriate forum and then click on the New Thread button.

3

I suppose that you should change the field separator.
If you post a sample data it would be easier.

4

Here is the data. This is the output of "df" command.

Filesystem 1K-blocks Used Available Use% Mounted on
/dev/mapper/VolGroup00-LogVol00
35772016 12316720 21608820 37% /
/dev/hda1 101086 13492 82375 15% /boot
tmpfs 451920 0 451920 0% /dev/shm
/dev/sda1 1013284 -7804 1021088 0% /media/LEXAR MEDIA

I have to extract the last field i.e mount point.
In the last line, the mount point has a space. so how to extract "/media/LEXAR MEDIA" from this line?

5

Instead of using the awk why can;t you use the cut command. pls fidn the below solution for your requirement.

$ echo "1 2 3 4 5 6 7 8 9"|cut -d" " -f6-

OUTPUT Will be
6 7 8 9

6

cut -d" " -f6- <your_data_file>

This will display contents from 6th column upto the end of line .

7

Something like this:

awk -F'% ' 'NR>1{print $NF}'
8

Thank u all for the replies.
Able to resolve the issue with `cut -d"%" -f2- | cut -d" " -f2-`